Conservative fields and potential functions
Problem 12.280 · medium
Show that \( \displaystyle \mathbf F = \left(y e^{x}\right)\mathbf i + \left(e^{x} + \cos{\left(y \right)}\right)\mathbf j \) is conservative, find a potential function \( \displaystyle f \), and evaluate \( \displaystyle \int_C \mathbf F \cdot d\mathbf r \) along any path from \( \displaystyle (0, -1) \) to \( \displaystyle (-1, 0) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(e^{x} + \cos{\left(y \right)}\right)\\\frac{\partial}{\partial y} y e^{x}\end{matrix}\right] = \left[\begin{matrix}e^{x}\\e^{x}\end{matrix}\right] \]∂Q/∂x = ∂P/∂y: F is conservative.✓ Proved
- Integrate the first component in x, then fix the 'constant' (a function of the other variables) by matching the other components.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(y e^{x} + \sin{\left(y \right)}\right)\\\frac{\partial}{\partial y} \left(y e^{x} + \sin{\left(y \right)}\right)\end{matrix}\right] = \left[\begin{matrix}y e^{x}\\e^{x} + \cos{\left(y \right)}\end{matrix}\right] \]f = y*exp(x) + sin(y) has gradient F.✓ Proved
- \[ 0 + \sin{\left(1 \right)} + 1 = \sin{\left(1 \right)} + 1 \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = y e^{x} + \sin{\left(y \right)} + C,\quad \int_C \mathbf F\cdot d\mathbf r = \sin{\left(1 \right)} + 1 \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the line integral computed numerically along two different paths |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The evaluation of the potential function at the endpoints is incorrect. f(0, -1) = -sin(1), so f(-1, 0) - f(0, -1) = -e^{-1} + sin(1). The solution incorrectly calculates f(0, -1) as 0 or ignores the exponential term.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-11 — The evaluation of the potential function at the endpoints is incorrect. f(0, -1) = -sin(1), so f(-1, 0) - f(0, -1) = -e^{-1} + sin(1). The solution incorrectly calculates f(0, -1) as 0 or ignores the exponential term.gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: fail (style) 2026-10-11 — [domain objection, downgraded to style] The potential function f(x,y) = y e^x + sin(y) is incorrect because its partial derivative with respect to y is e^x + cos(y), which matches Q, but its partial derivative with respect to x is y e^x, which matches P. Wait, let me re-check. P = y e^x, Q = e^x + cos(y). df/dx = y e^x (matches P). df/dy = e^x + cos(y) (matches Q). The potential function is actually correct. Let me re-read the solution steps. Step 1 checks conservativeness. Step 3 verifies the potential. Step 4 evaluates the integral. f(end) = f(-1, 0) = 0*e^{-1} + sin(0) = 0. f(start) = f(0, -1) = -1*e^0 + sin(-1) = -1 - sin(1). Integral = f(end) - f(start) = 0 - (-1 - sin(1)) = 1 + sin(1). The stated answer is sin(1) + 1. The calculation in step 4 is '0 + (sin(1) + 1)'. This implies f(start) was calculated as -(sin(1)+1). f(0, -1) = -1 - sin(1). So -f(start) = 1 + sin(1). The math is correct. Why did I think it was wrong? Let me look closer at Step 1. It computes Derivative(exp(x) + cos(y), x) which is exp(x). And Derivative(y*exp(x), y) which is exp(x). These are equal. So F is conservative. Step 3 checks the gradient of the proposed potential. It is correct. Step 4 calculates the line integral. f(-1,0) = 0. f(0,-1) = -1 - sin(1). Difference is 1 + sin(1). The solution says '0 + (sin(1) + 1)'. This is numerically correct. Is there a subtle error? The problem asks to 'find a potential function f'. The solution provides one. It asks to evaluate the integral. The solution does so. The sentences are: 'Integrate the first component in x...'. This is a valid method. 'f = ... has gradient F'. Valid. 'Integral = f(end) - f(start)'. Valid. Wait, look at Step 1 again. The matrix is [[dQ/dx], [dP/dy]]. dQ/dx = d/dx (e^x + cos y) = e^x. dP/dy = d/dy (y e^x) = e^x. They are equal. The text says '∂Q/∂x = ∂P/∂y: F is conservative.' This is the correct condition for a simply connected domain (which R^2 is). Everything seems correct. Let me double check the 'unchecked' tags. The prompt says 'Each line ... is either an EQUATION ... or a SENTENCE'. The user provided solution has equations marked 'unchecked'. The prompt says 'a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. But here they are marked 'unchecked'. However, the instructions say 'do not re-check the algebra of a line marked proved'. It doesn't explicitly say I must check unchecked lines, but usually 'unchecked' implies I should verify or that the CAS didn't run. But the prompt says 'Your job is to find what is wrong... Check that ... each conclusion follows from the lines before it'. If the algebra is wrong, it's an error. Let's re-verify the algebra just in case. Step 1: d/dx(e^x+cos y) = e^x. d/dy(y e^x) = e^x. Correct. Step 3: d/dx(y e^x + sin y) = y e^x. d/dy(y e^x + sin y) = e^x + cos y. Correct. Step 4: f(-1,0) = 0. f(0,-1) = -1 - sin(1). f(end)-f(start) = 1 + sin(1). Correct. Is there a sign error in the problem statement vs solution? Problem: F = (y e^x) i + (e^x + cos y) j. Solution uses same. Path from (0, -1) to (-1, 0). Solution uses same. Potential f = y e^x + sin y. Correct. Integral value 1 + sin(1). Correct. Why would this fail? Maybe the sentence in step 2 is too vague? 'Integrate the first component in x, then fix the constant...'. This is a standard procedure. Maybe the sentence in step 4 is misleading? '0 + (sin(1) + 1) = sin(1) + 1'. It doesn't explicitly state f(end) and f(start), but the result is correct. Let's look for a 'style' or 'misleading' issue. Or maybe I missed a detail. 'Show that F is conservative'. Step 1 does this. 'Find a potential function f'. Step 3 provides it. 'Evaluate integral'. Step 4 does this. Is the potential function unique? No, up to a constant. The solution gives one specific one. That is fine. Is the domain an issue? No, entire plane. Is the path issue? No, conservative field. I am struggling to find an error. Let me look at the JSON output format again. If it is sound, say pass. Is it sound? Yes. Wait, look at Step 1's text: '∂Q/∂x = ∂P/∂y: F is conservative.' This is a sufficient condition on a simply connected domain. The domain is R^2, which is simply connected. So the logic holds. Look at Step 4's text: '∫_C F·dr = f(end) − f(start), for any path.' This is the Fundamental Theorem of Line Integrals. It applies because F is conservative. The logic holds. I will mark it as pass.gpt-oss:20b: pass 2026-10-11
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/conservative_fields, checked 2026-10-11 with SymPy 1.14.0.